Let x, y, z denote the cost of 1 kg of Wheat, Sugar and Rice respectively.
2x + y = 70
x + z = 70
3x + 2y + z = 17
The given system can be written as
\( \begin{bmatrix} 2&1&0\\1&0&1\\3&2&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 70\\70\\170 \end{bmatrix}\) 
\(A X=B \Rightarrow X=A^{-1} B\)
Where \(A=\begin{bmatrix} 2&1&0\\1&0&1\\3&2&1 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix}B=\begin{bmatrix} 70\\70\\170 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 2&1&0\\1&0&1\\3&2&1 \end{vmatrix}=2\begin{vmatrix} 0&1\\2&1 \end{vmatrix}-1\begin{vmatrix} 1&1\\3&1 \end{vmatrix}+0\begin{vmatrix} 1&0\\3&2 \end{vmatrix} \)
\(=2(0-2)-1(1-3)+0=-4+2+0=-2\neq0\)
\(\therefore \) A^-1 exists 
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 2 & 2 \\ -1 & 2 & -1 \\ 1 & -2 & -1 \end{array}\right)\)
\(\operatorname{adj} A=\left(\begin{array}{ccc} -2 & -1 & 1 \\ 2 & 2 & -2 \\ 2 & -1 & -1 \end{array}\right)\)
\({A}^{-1}={{1}\over{|A|}}adj\ A={{-1}\over{2}}\begin{bmatrix} -2&-1&1\\2&2&-2\\2&-1&-1 \end{bmatrix}\)
\(X={A}^{-1}B={{-1}\over{2}}\begin{bmatrix} -2&-1&1\\2&2&-2\\2&-1&-1 \end{bmatrix}\begin{bmatrix} 70\\70\\170 \end{bmatrix}={{-1}\over{}2}\begin{bmatrix} -40\\-60\\-100 \end{bmatrix}=\begin{bmatrix} 20\\30\\50 \end{bmatrix}\)
\(\therefore\) Cost of 1kg of Wheat, Sugar and Rice are Rs. 20, Rs. 30 and Rs. 50 respectively.

a₁₁ = 30, a₁₂ = 40, x₁ = 120
a₂₁ = 20, a₂₂ = 10, x₂ = 60
\({b}_{11}={{{a}_{11}}\over{x_1}}={{30}\over{120}}={{1}\over{4}}\)
\({b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{40}\over{60}}={{2}\over{3}}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}={{20}\over{120}}={{1}\over{6}}\)
\({b}_{22}={{{a}_{22}}\over{{x}_{1}}}={{10}\over{60}}={{1}\over{6}}\)
The technology matrix is   B =  \(\begin{bmatrix}{{1}\over{4}}&{{2}\over{3}}\\ {{1}\over{6}}&{{1}\over{6}} \end{bmatrix}\)
I - B = \(\begin{bmatrix} 0&0\\0&1 \end{bmatrix}-\begin{bmatrix} {{1}\over{4}} &{{2}\over{3}}\\{{1}\over{6}}&{{1}\over{6}} \end{bmatrix}=\begin{bmatrix} {{3}\over{4}}&{{-2}\over{3}}\\ {{-1}\over{6}}&{{5}\over{6}} \end{bmatrix}\)
\(|I-B|=\frac{3}{4} \times \frac{5}{6}-\frac{2}{3} \times \frac{1}{6}=\frac{5}{8}-\frac{1}{9}=\frac{37}{72}=0\)
Since diagonals of I - B are positive and | I - B | is positive, the system is viable
\({(I-B)}^{-1}={{1}\over{|I-B|}}adj\ (I-B)={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\)
X = (I - B)^-1 D where D = \(\begin{bmatrix} 80\\40 \end{bmatrix}\)
\(={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\begin{bmatrix} 80 \\ 40 \end{bmatrix}\)
\(=\frac{72}{37}\left(\begin{array}{cc} 66.67 & +26.67 \\ 13.33 & +30 \end{array}\right)=\frac{72}{37}\left(\begin{array}{l} 93.34 \\ 43.33 \end{array}\right)\)
\(=\left(\begin{array}{c} 181.63 \\ 84.32 \end{array}\right)\)
The output for production section X and Y are 181.63 and 84.32 respectively.

Given A \(=\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=(4+7)-1(3-7)+1(-3-4)\)
\(=11+4-7=8\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 11 & 4 & -7 \\ -2 & 0 & +2 \\ 3 & -4 & 1 \end{array}\right)\)
\(\operatorname{adj} A=\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(\mathrm{A}(\operatorname{adj} A)=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right)\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 11+4-7 & -2+0+2 & 3-4+1 \\ 33+16-49 & -6+0+14 & 9-16+7 \\ 11-4-7 & -2+0+2 & 3+4+1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right)=8\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I_3\)
\((adj\ A)\ A=\begin{bmatrix}11&-2&3\\4&0&-4\\-7&2&1\end{bmatrix}\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=\begin{bmatrix} 11-6+3&11-8-3&11-14+3\\4+0-4&4+0+4&4+0-4\\-7+6+1&-7+8-1&-7+14+1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8 \end{bmatrix}\) ...(2)
\(|A|.{I}_{3}=8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8\end{bmatrix}\)  ...(3)
From (1), (2) and (3)
A (adj A) = (adj A) A = |A|I₃.

LHS = \(\begin{vmatrix}-a^2&ab&ac\\ab&-b^2&bc\\ac&bc&-c^2 \end{vmatrix}\)
Taking a, b, c common from R₁, R₂ and R₃ respectively we get.
LHS = \(abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c \end{vmatrix}\)
Again taking a, b, c common from C₁ C₂ and C₃ respectively.
LHS = \(a^2b^2c^2\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1 \end{vmatrix}\)
\(R_2 \rightarrow R_2+R_1, R_3 \rightarrow R_3+R_1\)
\(=a^2b^2c^2\begin{vmatrix}0&0&2\\1&-1&1\\1&1&-1 \end{vmatrix}\)
\(=a^2 b^2 c^2[-1(0-4)]=4 a^2 b^2 c^2\)
= RHS Hence proved.

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)